Uniform Convergence of a Parameter Series on [0,1)
Consider the series S(x)=∑_{n=1}∞ x^n/(1+x^n), defined for x∈[0,1). A homework asks two questions: (i) is the series uniformly convergent on [0,a] for each fixed 0<a<1? (ii) is it uniformly convergent on [0,1)? A student writes: 'The Weierstrass M-test with M_n=1 fails because ∑1 diverges; therefore the series is not uniformly convergent on either interval. Nevertheless S is continuous on [0,1) because each term is continuous and the series converges pointwise.' The TA is unsure, because failure of the M-test does not prove failure of uniform convergence. The instructor asks you to settle both uniform-convergence questions, assess the continuity claim, and give a correct reason for continuity if it is true. Use the Cauchy criterion or tail estimates, not just the M-test.
1. Prove or disprove uniform convergence on [0,a] for fixed 0<a<1.
Read the worked answer
The series is uniformly convergent on [0,a]. For 0≤x≤a<1, the denominator satisfies 1+x^n≥1, so 0≤x^n/(1+x^n)≤x^n≤a^n. Since 0<a<1, the geometric series ∑a^n converges. The Weierstrass M-test with M_n=a^n therefore applies and gives uniform convergence on [0,a].
The student's use of M_n=1 is a failed test, not a proof of nonuniform convergence. A failed M-test can mean only that this particular dominating series is too crude. Here replacing the crude bound 1 by the x-dependent bound a^n captures the decay that is uniform on the compact subinterval [0,a].
2. Show that the series is not uniformly convergent on [0,1).
Read the worked answer
Use the Cauchy criterion. Fix N and set x_N=(1/2)^{1/(2N)}. Then x_N<1 and x_N^{2N}=1/2. For every n with N+1≤n≤2N, we have x_N^n≥x_N^{2N}=1/2. Hence x_N^n/(1+x_N^n)≥(1/2)/(1+1)=1/4, because the denominator is at most 2. Summing over the N terms n=N+1,…,2N gives a tail of size at least N/4.
Therefore, for ε0=1/8, no N can make every tail from N+1 to 2N smaller than ε0 uniformly in x: at x=x_N the tail is at least 1/4. The Cauchy criterion for uniform convergence fails, so the series is not uniformly convergent on [0,1). The obstruction is the behavior near x=1, where the terms approach 1/2 and the series behaves like a divergent harmonic-type sum.
3. Is S continuous on [0,1)? Assess the student's reason.
Read the worked answer
Yes, S is continuous on [0,1), but the student's stated reason is invalid. Pointwise convergence of continuous functions does not preserve continuity in general: f_n(x)=x^n on [0,1] converges pointwise to a function that is 0 on [0,1) and 1 at 1.
The correct argument uses uniform convergence on compact subintervals. Fix x0∈[0,1). Choose a with x0<a<1. On [0,a], the series converges uniformly by the M-test with a^n, and every term x^n/(1+x^n) is continuous. A uniformly convergent series of continuous functions has a continuous sum, so S is continuous on [0,a]. Since x0 was arbitrary and every point of [0,1) lies in some such [0,a], S is continuous on all of [0,1). Uniform convergence on the whole half-open interval is not needed; local uniform convergence on compact subintervals suffices.